Determining a limit without L'Hospital's Rule

Question

I'm trying to solve the limit $$\lim_{h\to0}\frac{\sin(\pi+h)}{h(\pi+h)}$$ without using L'Hospital's rule. It's part of a problem for which I am trying to prove that $$\lim_{h\to0}\frac{\frac{\sin(\pi+h)}{|\pi+h|}-\frac{h}{\pi}}{|h|}$$ exists. After a bit of simplifying and assuming $h>0$ I have arrived at the first expression.

Answer 1

\begin{align} \lim_{h\to 0}\frac{\sin(\pi+h)}{h(\pi+h)}&=-\lim_{h\to 0}\frac{\sin h}{h} \cdot\frac1{\pi+h}\\ &=-\lim_{h\to 0}\frac1{\pi+h}\\ &=-\frac1\pi \end{align}

Answer 2

With high-school tools: $$\frac{\sin(\pi+h)}{h(\pi+h)}=-\underbrace{\frac{\sinh}{h}}_{\substack{\downarrow\\1}}\underbrace{\frac1{\pi+h}}_{\substack{ \downarrow\\ 1/\pi}}.$$

Answer 3

Note that $\sin(\pi + h) = -\sin(h)$. We can use the Taylor series of $\sin$ to arrive at

$\lim_{h \to 0} \frac{-h + h^3/3! - h^5/5! + ...}{h(\pi + h)} = \lim_{h \to 0} \frac{-1 + h^2/3! - h^4/5! + ...}{\pi + h} = -\frac{1}{\pi}$.

Answer 4

Note that $$\sin (h+\pi )=-\sin(h)$$

Thus the limit is $$ \lim _{h\to 0}-((\sin h)/h )(1/\pi )=(-1/\pi)$$

Answer 5

One possible approach consists in using the following results \begin{align*} \begin{cases} \displaystyle\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1\\\\ \displaystyle \sin(\pi + h) = -\sin(h) \end{cases} \end{align*}

At your case, we have \begin{align*} \lim_{h\rightarrow 0}\frac{\sin(\pi + h)}{h(\pi + h)} = -\lim_{h\rightarrow 0}\left[\frac{\sin(h)}{h}\times\frac{1}{\pi + h}\right] = -\frac{1}{\pi} \end{align*}

Answer 6

You're trying to evaluate $\lim_{h\to0}\frac{-\sin h}{h}\cdot\frac{1}{\pi+h}=-\frac{1}{\pi}$, so that you can then prove the existence of$$\lim_{h\to0}\frac{\frac{\sin(\pi+h)}{|\pi+h|}-\frac{h}{\pi}}{|h|}.$$Since the latter two-sided limit is of an odd function of $h$, to exist we need it to be $0$. As $h\to0$,$$\frac{\sin(\pi+h)}{|\pi+h|}-\frac{h}{\pi}=-\frac{1}{\pi}\left(\frac{\sin h}{1+h/\pi}+h\right)\sim\frac{-2h}{\pi},$$so division by $|h|$ gives $-\frac{2h}{\pi|h|}$ as the asymptotic behaviour. Thus the limit you wanted to exist doesn't.

Answer 7

Recall that $\sin(\pi+h)=-\sin h$ then

$$\frac{\sin(\pi+h)}{h(\pi+h)}=\frac{-\sin(h)}{h}\cdot\frac{1}{\pi+h}\to-1\cdot \frac1\pi=-\frac1\pi$$